已知sina sinb=1 2,cosa-cosb=1 3,求cos(a b)

2sinAsinB=cos(A-B)-cos(A+B)=cos(A-B)+cosC=1+cosC所以cos(A-B)=1,A=B,三角形ABC是等腰三角形.

(a^3+b^3+c^3)/(a+b+c)=c^2a^3+b^3+c^3=ac^2+bc^2+c^3a^3+b^3-ac^2-bc^2=0(a+b)(a^2-ab+b^2)-c^2(a+b)=0(a+

由sin^2A+sin^2B-sinAsinB=sin^2C由正弦定理sinA=a/2R,sinB=b/2R,sinC=c/2R则(a/2R)^2+(b/2R)^2-(a/2R)(b/2R)=(c/2

1）移项：cosAcosB-sinAsinB=cos(A+B)=0所以A+B=π/2即C=π/2所以为直角三角形2)因为tanB=tanC>1所以90°>B>45°,90°>C>45°所以A

原式可化为a^2+b^2-c^2=ab也即是a^2+b^2-c^2/2ab=1/2也即是cosC=1/2所以C=60°联立2sinC=sinA+sinB可得等边三角形

题目应该是a²+b²=c²+ab,sinA•sinB=3/4吧△ABC是等边三角形余弦定理：cosC=(a²+b²-c²)/2a

在△ABC中,已知2sinAsinB=cosC,试判断△ABC的形状由2sinAsinB=cosC,得cos(A-B)-cos(A+B)=cosCA+B=180°-C,代入上式得cos(A-B)+co

已知(a+b+c)(a+b-c)=3ab(a+b)²-c²=3aba²+b²-c²=abc²=a²+b²+abC=60A

a^2+b^2=c^2+abc^2=a^2+b^2-ab余弦定理,c^2=a^2+b^2-2*a*b*cosCcosC=1/2,C=60°,A+B=120°cos(A+B)=cosAcosB-sinA

由正弦定理变形为a²+b²+ab=c²,a²+b²-c²=-ab∴由余弦定理,cosC=(a²+b²-c²)/

cos²0.5C=cos²0.5(180-(A+B))=cos²[90-0.5(A+B)]=sin²0.5(A+B)=0.5-0.5cos(A+B)sinAsi

【ps：一般性题目就像下面这么出的,按你原题做肯定不是特殊三角形,所以我估计是题目出错了,等星期一问问老师吧】这道题目是否出错?应该是(a³+b³-c³)/(a+b-c)

2sinAsinB=cos(A-B)-cos(A+B)=cos(A-B)+cosC=1+cosC所以cos(A-B)=1,A=B,三角形ABC是等腰三角形.

等腰三角形证明：sinAsinB=cos²（C/2）=(cosC+1)/21+cosC=2sinAsinB=cos(A-B)-cos(A+B)cos(A+B)=cos(180-C)=-cos

是sin²A+sin²B-√2sinAsinB吧.由余弦定理得cosC=(a²+b²-c²)/(2ab)由正弦定理得(sin²A+sin&#

三角形ABC中1+cosC=1+cos(180-A-B)=!-cos(A+B)=1-cosAcosB+sinAsinB=2sinAsinB所以sinAsinB+cosAcosB=1即cos(A-B)=

等腰证明：sinAsinB=cos²（C/2）=(cosC+1)/21+cosC=2sinAsinB=cos(A-B)-cos(A+B)cos(A+B)=cos(180-C)=-cosC所以

sinasinb=1∴sina=sinb=1或者sina=sinb=-1.∴a=b=π/2+2kπ或者a=b=-π/2+2kπ,k∈Z∴cos((a+b)/2)=0

根据cosC=2cos^2(C/2)-1sinAsinB=0.5*(cosC+1)sinAsinB=0.5cos(pi-A-B)+0.5sinAsinB=-0.5cos(A+B)+0.5sinAsin

是直角三角形由正弦定理得(a+b)/a==sinB/(sinB-sinA)=b/(b-a)所以b^2-a^2=ab又因为2sinAsinB=2sin^2C,得ab=c^2所以有b^2-a^2=c^2也